Characteristic polynomial of a linear operator in matrix space

For $$A \in M_{n\times n}(F)=V$$, we define a linear operator $$T$$ such that $$T(X)=AX$$ for $$X\in V$$.

I need to show that the characteristic polynomial of $$T$$ is $$(f_A(t))^n$$, where $$f_A(t)$$ is the characteristic polynomial of $$A$$.

My latest approach has been to define $$W$$, the $$T$$-cyclic subspace of $$V$$ generated by some $$X$$. Then,

$$\beta = \{X,\,T(X),\,T^2(X),\,\dots,\,T^{k-1}(X)\}$$

is an ordered basis for $$W$$, where $$k=dim(W)$$. From here, I'm thinking I need to extend the basis to $$n$$ so that it's a basis for $$V$$ and then show somehow that, since $$T^n(X)=A^nX$$, my desired result falls out.

Am I on the right track? Any pushes in the right direction that could help me put the pieces together? Or is there a totally different way to solve this that has eluded me?

• Another route I have found to investigate is to start by using the fact that the characteristic polynomial of any $A\in M_{n\times n}(F)$ is equal to $(-1)^n(a^nt^n + a_{n-1}t^{n-1} + \dots + a_1t + a_0$). Nov 15, 2018 at 22:25
• Nevermind, I've managed to solve it. See my answer below. Nov 16, 2018 at 8:48

Extending the ordered basis so that $$k=n$$, we have

$$\beta = \{X, AX, A^2X,...,A^{n-1}X\}$$

as a basis for $$V$$. Now we simply express $$T$$ as $$[T]_{\beta}$$, the matrix form of the operator, and take the characteristic polynomial of the result. Note that:

$$T(X)=AX$$

$$T(AX)=A^2X$$

$$\dots$$

$$T(A^{n-1}X)=A^nX$$

and thus

$$[T]_{\beta}=AI_n$$

which has the characteristic polynomial given by $$(f_A(t))^n$$, as needed.